Integrable Systems in n-dimensional Riemannian Geometry

Jan A. Sanders Faculty of Sciences Vrije Universiteit

1081 HV Amsterdam The Netherlands

Jing Ping Wang ^{∗} Department of Mathematics Brock University St. Catharines, Ontario L2S 3A1 Canada

February 16, 2006

Abstract

In this paper we show that if one writes down the structure equations for the evolution of a curve embedded in an n-dimensional Riemannian manifold with constant curvature this leads to a symplectic, a Hamiltonian and an hereditary operator. This gives us a natural connection between finite dimensional geometry, infinite dimensional geometry and integrable systems. Moreover one finds a Lax pair in 𝔬_{n+1} with the vector modified Korteweg-De Vries equation (vmKDV)

u_{t} = u_{xxx} + \frac{3}{2} ||u||^{2}u_{x}

as integrability condition. We indicate that other integrable vector evolution equations can be found by using a different Ansatz on the form of the Lax pair. We obtain these results by using the natural or parallel frame and we show how this can be gauged by a generalized Hasimoto transformation to the (usual) Frenêt frame. If one chooses the curvature to be zero, as is usual in the context of integrable systems, then one loses information unless one works in the natural frame.

1 Introduction

The study of the relationship between finite-dimensional differential geometry and partial differential equations which later came to be known as integrable systems, started in the 19th century. Liouville found and solved the equation describing minimal surfaces in 3-dimensional Euclidean space [Lio53]. Bianchi solved the general Goursat problem for the sine-Gordon equation, which arises in the theory of pseudospherical surfaces [Bia53, Bia27].

∗We thank N WO (Netherlands Organization for Scientific Research) for financial support and the Newton Institute for its hospitability and support.

Much later Hasimoto [Has72] found the relation between the equations for curvature and torsion of vortex filament flow and the nonlinear Schrödinger equation, which led to many new developments, cf. [MW83, LP91, DS94, LP96, YS98, Cal00].

The similarity between the structure equations for connections in differential geometry and Lax pair equations for integrable systems has intrigued many researchers from both fields of mathematics. We refer the interested reader to the following books: [BE00, Gue97, Hél01, Hél02, RS02]. A good introductory review is [Pal97].

Recently, we showed in [MBSW02] that if a flow of curves in a 3–dimensional Riemannian manifold with constant curvature κ follows an arc–length preserving geometric evolution, the evolution of its curvature and torsion is always a Hamiltonian flow with respect to the pencil E + D + κC, where E, D and C are compatible Hamiltonian structures. However, the close geometric relationship remains here: the triplet is obtained solely from the intrinsic geometry of curves on 3–dimensional Riemannian manifolds with constant curvature.

Once one has two compatible Hamiltonian operators, one can construct an hereditary operator, from which a hierarchy of integrable equations can be computed. They are all Hamiltonian with respect to two different Hamiltonian operators, that is, biHamiltonian as defined in [Mag78]. Thus Poisson geometry is very important in the study of integrable systems, cf. [Olv93, Dor93].

The goal of the present paper is to generalize this analysis to arbitrary dimension and see how much of the infinite dimensional geometric structure is still present in this case.

The Hasimoto transformation is a Miura transformation, which is induced by a gauge transformation from the Frenêt frame (the obvious frame to choose from the traditional differential geometry point of view) to the parallel or natural frame, cf. [DS94]. On the basis of loc.cit., Langer and Perline [LP00] advocated the use of the natural frame in the n-dimensional situation in the context of vortex-filament flow.

We show in this paper that the appreciation of the exact relationship between the underlying finite dimensional geometry and the infinite dimensional geometry has been complicated by the use of the Frenêt frame. In fact, the operators that lead to biHamiltonian systems naturally come out of the computation of the structure equations. This fact is true in general, but only when one uses the parallel or natural frame does this conclusion come out automatically. This will be the first ingredient of our approach.

To relate our results to the classical situation in terms of the curvatures of the evolving curve in the Frenêt frame we can of course say that this relation is obvious since there must exist a gauge transform between the two connections, defined by the Cartan matrices specifying our frame. Nevertheless, we give its explicit construction, and produce a generalized Hasimoto transformation in arbitrary dimension, with complete proof, as announced in [Wan03]. This insures that anything that can be formulated abstractly can also be checked by direct (though complicated) computations. Since the equations are derived from the computation of the curvature tensor which behaves very nicely under transformations, we feel that these direct calculations are the very last thing to try. In this kind of problem the abstract point of view seems to be much more effective computationally. With hindsight one might say that Hasimoto was the first to exploit the natural frame to show the integrability of the equations for the curvature and torsion of a curve embedded in a three-dimensional Riemannian manifold.

Here we should mention the fact that the generalization of the Hasimoto also plays a role in [LP00], but it is in a different direction; the focus is on the complex structure and this is generalized to the 2n-dimensional situation.

A second ingredient in our approach is the fact that we assume the Riemannian manifold to have (nonzero) constant curvature. If one takes the curvature equal to zero and one uses the natural frame, then one can still recover the symplectic and cosymplectic operators which give rise to a biHamiltonian system. But if one uses another frame, then information gets lost and one is faced with the difficult task of recreating this information in order to get the necessary operators and Poisson geometry. In our opinion, this has been a major stumbling block in the analysis of the relation between high-dimensional geometry and integrability.

A third ingredient is the study of symmetric spaces. The relation between symmetric spaces and integrability was made explicit in the work of Fordy and others [AF87, FK83], There it is shown how to construct a Lax pair using the structure of the Lie algebra. This construction does not work in the semidirect product 𝔢𝔲𝔠_{n}, but there we use the fact that we can map the elements in the n-dimensional euclidean algebra to the n + 1-dimensional orthogonal algebra. This leads immediately to the construction of Lax pairs, combining the methods in [TT01] and [LP00].

Putting together these three elements, natural frame, constant curvature and extension to a symmetric semisimple Lie algebra, gives us complete control over all the integrability issues one might like to raise. We have tried to formulate things in such a way that it is clear what to do in other geometries. The theory, however, is not yet strong enough to guarantee success in other geometries, so this will remain an area of future research.

If we try to understand the success of the natural frame in this context, then it seems that the key is the natural identification of the Lie algebra 𝔢𝔲𝔠_{n}/𝔬_{n} with the adjoint orbit of 𝔬_{n}^{∗}. In the natural frame, this identification is simple to see and can be performed by putting a complex structure on 𝔬_{n+1}. To give a geometrical explanation of this, one may have to formulate everything in terms of Poisson reduction of the Kac-Moody bracket of SO(n), but we make no attempt to do so here. The main point we will make in this paper is that the whole construction can be explicitly computed. Once this is done, one can then choose a more conceptional approach later on. We have indicated the general character of the approach where appropriate.

The paper is organized as follows. In section 2 we derive the structure equations, using Cartan’s moving frame method, and find that the equation is of the type

u_{t} = Φ_{1}h + κΦ_{2}h, Φ_{2} = Id_{n−1}.

In section 4 we show that Φ_{1} is an hereditary operator, and we draw the conclusion that if we had been computing in any other frame, resulting in the equation

ū_{t} = Φ̄_{1}h̄ + κΦ̄_{2}h̄,  (1)

then Φ̄_{1}Φ̄_{2}^{−1} is also hereditary.

We then proceed, by taking h = u_{1}, the first x-derivative of u, to derive a vector mKDV equation. In [DS94] it was predicted (or derived by general considerations) that the equation should be a vmKDV equation, but there are two versions of vmKDV equation, cf. [SW01b], and the prediction does not say which version it should be. That question is now settled.

In section 3 we state explicitly the formula for the generalized Hasimoto transformation, which transforms the Frenêt frame into the natural frame and in Appendix B we prove this.

We conclude in section 5 by constructing a Lax pair which has this equation as its integrability condition. This is done as follows. The geometric problem is characterized by two Lie algebras, 𝔬_{n} and 𝔢𝔲𝔠_{n}. The usual Lie algebraic construction of Lax pairs starts with a symmetric Lie algebra 𝔥^{0} + 𝔥^{1}, with

[𝔥^{0}, 𝔥^{0}] ⊂ 𝔥^{0}, [𝔥^{0}, 𝔥^{1}] ⊂ 𝔥^{1}, [𝔥^{1}, 𝔥^{1}] ⊂ 𝔥^{0}.  (2)

If we think of 𝔬_{n} as 𝔥^{0}, then we cannot take 𝔢𝔲𝔠_{n} as 𝔥^{0} + 𝔥^{1}, since 𝔥^{1} is then abelian. Although it trivially obeys all the inclusions, this triviality kills all the actions we need to construct the Lax pair. But if we take 𝔬_{n+1} = 𝔥^{0} + 𝔥^{1}, then things fit together nicely and we can simply copy the construction in [TT01]. Of course, we can identify 𝔥^{1} with 𝔢𝔲𝔠_{n}/𝔬_{n} as vectorspaces, and this should be the main consideration (apart from the requirements in (2)) in other geometries when one tries to find such a symmetric extension of the geometrically given 𝔥^{0}.

Acknowledgment The authors thank Gloria Marí Beffa for numerous discussions, which both influenced the questions that we asked ourselves as well as some the methods we used. The work was mainly carried out at the Isaac Newton Institute for Mathematical Sciences during the Integrable Systems programme in 2001 and we thank N WO (Netherlands Organization for Scientific Research) for financial support and the Newton Institute for its hospitability and support. Thanks go to Hermann Flaschka for reminding us of [TT01] and to Sasha Mikhailov for his help with Lax pairs.

2 Moving Frame Method

Our approach, although restricted to the Riemannian case, is also suitable for arbitrary geometries. As a general reference to the formulation of geometry in terms of Lie algebras, Lie groups and connections we refer to [Sha97].

We consider a curve (denoted by e_{0} if we want to stress its relation to the other vectors in the moving frame, or γ, if we just want to concentrate on the curve itself), parametrized by arclength x and evolving geometrically in time t, which is imbedded in a Riemannian manifold M. We choose a frame e_{1}, ⋯, e_{n} as a basis for T_{e_{0}}M, and a dual basis of one forms τ_{1}, ⋯, τ_{n} ∈ ⋀^{1} M, so that τ_{i}(e_{j}) = δ_{i}^{j}. Choosing a frame is equivalent to choosing an element in

g ∈ G = Euc(n, ℝ) = O(n, ℝ) ⋉ ℝ^{n}. We write g = ( \begin{matrix}1 & e_{0} \\ 0 & e \end{matrix} ). Let Γ(TM) be

the space of smooth sections, that is, maps σ : M → TM such that πσ is the identity map on M, where π is the natual projection of the tangent space on its base M. We consider the e_{i} as sections in TM, at least locally, that is, elements in Γ(TM), varying e_{0}. We assume that there is a Lie algebra 𝔤 and a subalgebra 𝔥 such that T_{e_{0}}M ⋍ 𝔤/𝔥 as vectorspaces.

We now define a connection d : ⋀^{p} M⊗Γ(TM) → ⋀^{p+1} M⊗Γ(TM) as follows. Let ω ∈ ⋀^{p} M. Then

dω ⊗ σ = dω ⊗ σ + (−1)^{p}ω ⊗ dσ.

Notice that we use d as a cochain map from ⋀^{p} M to ⋀^{p+1} M, p ∈ ℕ in the ordinary de Rham complex. We extend the connection to act on e_{0} as follows.

de_{0} = ∑_{i} τ_{i} ⊗ e_{i}, τ_{i} ∈ ⋀^{1} M

with τ = (τ_{1}, ⋯, τ_{n}) ∈ C^{1}(𝔤/𝔥, 𝔤/𝔥) a differential 1–form. We now suppose that de_{i} = ∑_{j} ω_{ij} ⊗ e_{j}, with ω_{ij} ∈ ⋀^{1} M, that is, ω ∈ C^{1}(𝔤/𝔥, 𝔥). This connection naturally extends to a connection d : ⋀^{p} M ⊗ Γ(G) → ⋀^{p+1} M ⊗ Γ(G). We may write our basic structure equations as

dg = Ag = ( \begin{matrix}0 & τ \\ 0 & ω \end{matrix} ) g  (3)

where A is the vector potential or Cartan matrix. We remark that dτ(X, Y) = Xτ(Y) − Y τ(X) − τ([X, Y]). Differentiating both sides of (3), we have

d^{2}g = ( \begin{matrix}0 & dτ \\ 0 & dω \end{matrix} ) g − ( \begin{matrix}0 & τ \\ 0 & ω \end{matrix} ) ( \begin{matrix}0 & de_{0} \\ 0 & de \end{matrix} )
= ( \begin{matrix}0 & dτ \\ 0 & dω \end{matrix} ) g − ( \begin{matrix}0 & τ \\ 0 & ω \end{matrix} ) ( \begin{matrix}0 & τ \\ 0 & ω \end{matrix} ) g
= ( \begin{matrix}0 & dτ − τ ∧ ω \\ 0 & dω − ω ∧ ω \end{matrix} ) g = Fg

We draw the conclusion that

d^{2} = F = ( \begin{matrix}0 & T \\ 0 & Ω \end{matrix} ) = ( \begin{matrix}0 & dτ − τ ∧ ω \\ 0 & dω − ω ∧ ω \end{matrix} ) ,

where F ∈ C^{2}(𝔤/𝔥, 𝔤), of which the 2–form Ω ∈ C^{2}(𝔤/𝔥, 𝔥) is called the curvature, the 2–form T is called torsion and ω ∧ ω is often denoted by \frac{1}{2}[ω, ω].

We are now in a position to compute a few things explicitly. Let X = e_{0∗} \frac{∂}{∂x} = D_{x} and Y = e_{0∗} \frac{∂}{∂t} = D_{t}, that is, the induced vectorfields tangent to the imbedded curve of \frac{∂}{∂x} and \frac{∂}{∂t}, respectively.

Then, using the fact that [X, Y] = 0, we obtain

F(X, Y) =
= ( \begin{matrix}0 & T(X, Y) \\ 0 & Ω(X, Y) \end{matrix} ) = ( \begin{matrix}0 & dτ(X, Y) − τ(X) ∧ ω(Y) \\ 0 & dω(X, Y) − ω(X) ∧ ω(Y) \end{matrix} )
= ( \begin{matrix}0 & Xτ(Y) − Y τ(X) − τ(X) ∧ ω(Y) \\ 0 & Xω(Y) − Y ω(X) − ω(X) ∧ ω(Y) \end{matrix} )
= ( \begin{matrix}0 & D_{x}τ(Y) − D_{t}τ(X) − τ(X) ∧ ω(Y) \\ 0 & D_{x}ω(Y) − D_{t}ω(X) − ω(X) ∧ ω(Y) \end{matrix} )  (4)
= [D_{x} + ( \begin{matrix}0 & τ(X) \\ 0 & ω(X) \end{matrix} ) , D_{t} + ( \begin{matrix}0 & τ(Y) \\ 0 & ω(Y) \end{matrix} )]  (5)

3 Hasimoto transformation

In 1975, Bishop discovered the same transformation as Hasimoto when he studied the relations between two different frames to frame a curve in 3-dimensional Euclidean space, namely the Frenêt frame and parallel (or natural) frame, cf. [Bis75]. More explicitly, let the orthonormal basis {T, N, B} along the curve be the Frenêt frame, that is,

( \begin{matrix}T \\ N \\ B \end{matrix} )_{x} = ( \begin{matrix}0 & κ & 0 \\ −κ & 0 & τ \\ 0 & −τ & 0 \end{matrix} ) ( \begin{matrix}T \\ N \\ B \end{matrix} ) .

The matrix in this equation is called the Cartan matrix. Now we introduce the following new basis

( \begin{matrix}T \\ N^{1} \\ N^{2} \end{matrix} ) = ( \begin{matrix}1 & 0 & 0 \\ 0 & cos θ & − sin θ \\ 0 & sin θ & cos θ \end{matrix} ) ( \begin{matrix}T \\ N \\ B \end{matrix} ) , θ = ∫ τdx .  (6)

Its frame equation is

( \begin{matrix}T \\ N^{1} \\ N^{2} \end{matrix} )_{x} = ( \begin{matrix}0 & κ cos θ & κ sin θ \\ −κ cos θ & 0 & 0 \\ −κ sin θ & 0 & 0 \end{matrix} ) ( \begin{matrix}T \\ N^{1} \\ N^{2} \end{matrix} ) .

We call the basis {T, N^{1}, N^{2}} the parallel frame. This geometric meaning of Hasimoto transformation was also pointed out in [DS94], where the authors studied the connection between the differential geometry and integrability.

In this section, we give explicit formula of such a transformation in arbitrary dimension n, which has the exact same geometric meaning as the Hasimoto transformation. Therefore, we call it generalized Hasimoto transformation. The existence of such a transformation is clear from the geometric point of view, and was mentioned and implicitly used in several papers such as [Bis75] and [LP00].

First we give some notation. Denote the Cartan matrix of the Frenêt frame ē by ω̄(D_{x}) ( ē_{x} = ω̄(D_{x})ē ) and that of the parallel frame e by ω(D_{x}) ( e_{x} = ω(D_{x})e ), i.e.,

ω̄(D_{x}) = ( \begin{matrix}0 & ū^{(1)} & 0 & ⋯ & 0 \\ −ū^{(1)} & 0 & ū^{(2)} & ⋯ & 0 \\ ⋮ & ... & ... & ... & ⋮ \\ 0 & −ū^{(n−3)} & 0 & ū^{(n−2)} & 0 \\ 0 & 0 & −ū^{(n−2)} & 0 & ū^{(n−1)} \\ 0 & 0 & 0 & −ū^{(n−1)} & 0 \end{matrix} )

and, letting u = (u^{(1)}, u^{(2)}, ⋯, u^{(n−1)})^{⊺},

ω(D_{x}) = ( \begin{matrix}0 & u^{⊺} \\ −u & 0 \end{matrix} ) .

The orthogonal matrix that keeps the first row and rotates the i-th and j-th row with the angle θ_{ij} is denoted by R_{ij}, where 2 ≤ i < j ≤ n. For example, when n = 3, the orthogonal matrix

R_{23} = ( \begin{matrix}1 & 0 & 0 \\ 0 & cos θ_{23} & sin θ_{23} \\ 0 & − sin θ_{23} & cos θ_{23} \end{matrix} )

is the transformation between two frames, i.e., ē = R_{23}e, comparing to (6). Let us compute the conditions such that

ω̄(D_{x})R_{23} − \frac{∂R_{23}}{∂x} = R_{23}ω(D_{x}).

Then we have ū^{(2)} = D_{x}θ_{23}, u^{(1)} = ū^{(1)} cos θ_{23} and u^{(2)} = ū^{(1)} sin θ_{23}. This is the famous Hasimoto transformation, i.e., ϕ = u^{(1)}+iu^{(2)} = ū^{(1)} exp(i ∫ ū^{(2)}dx).

Therefore, the generalized Hasimoto transformation, which by definition transforms the Frenêt frame into the natural frame, can be found by computing the orthogonal matrix, which gauges the Cartan matrix of the Frenêt frame into that of the parallel frame in n-dimensional space.

Theorem 1. Let n ≥ 3. The orthogonal matrix R gauges the standard Frenêt frame into the natural frame, that is,

ω̄(D_{x})R − D_{x}R = Rω(D_{x}),

where R = R_{n−1,n} ⋯ R_{3,n} ⋯ R_{34}R_{2,n} ⋯ R_{24}R_{23}. This leads to the components of u satisfying the Euler transformation

u^{(1)} = ū^{(1)} cos θ_{23} cos θ_{24} ⋯ cos θ_{2n},

u^{(2)} = ū^{(1)} sin θ_{23} cos θ_{24} ⋯ cos θ_{2n},
u^{(3)} = ū^{(1)} sin θ_{24} ⋯ cos θ_{2n},
· · · · · ·
· · · · · ·
u^{(n−2)} = ū^{(1)} sin θ_{2,n−1} cos θ_{2n},
u^{(n−1)} = ū^{(1)} sin θ_{2n},

and the curvatures in the standard Frenêt frame satisfying

ū^{(i)} = ∏_{j=i+2}^{n} \frac{cos θ_{i,j}}{cos θ_{i+1,j}} A_{i}, i = 2, 3, ⋯, n − 1,  (7)

where the A_{i} are generated by the recursive relation

A_{i} = D_{x}θ_{i,i+1} + sin θ_{i−1,i+1}A_{i−1}, A_{1} = 0, 2 ≤ i ≤ n − 1.  (8)

There are \frac{1}{2}(n − 2)(n − 3) constraints on the rotating angles:

D_{x}θ_{ij} = ∏_{l=i+2}^{j} \frac{cos θ_{il}}{cos θ_{i+1,l}} sin θ_{i+1,j}A_{i} − ∏_{l=i+1}^{j−1} \frac{cos θ_{i−1,l}}{cos θ_{i,l}} sin θ_{i−1,j}A_{i−1}, (9)

where 4 ≤ j ≤ n, 2 ≤ i ≤ j − 2. Proof. It is obvious that R = ( \begin{matrix}1 & 0 \\ 0 & T \end{matrix} ), where T is an orthogonal (n − 1) × (n − 1)-matrix and the first row of T equals is \frac{u^{⊺}}{ū^{(1)}} , that is,

(cos θ_{23} cos θ_{24} ⋯ cos θ_{2n}, sin θ_{23} cos θ_{24} ⋯ cos θ_{2n}, ⋯, sin θ_{2n}) . This implies that v̄^{⊺}T = u^{⊺}, where v̄^{⊺} = (ū^{(1)}, 0, ⋯, 0). Since T T^{⊺} = I, we have T u = v̄. Rewrite ω̄(D_{x}) = ( \begin{matrix}0 & v̄^{⊺} \\ −v̄ & \underline{ω} \end{matrix} ), and compute

ω̄(D_{x})R − D_{x}R − Rω(D_{x})

= ( \begin{matrix}0 & v̄^{⊺} \\ −v̄ & \underline{ω} \end{matrix} ) ( \begin{matrix}1 & 0 \\ 0 & T \end{matrix} ) − ( \begin{matrix}0 & 0 \\ 0 & D_{x}T \end{matrix} ) − ( \begin{matrix}1 & 0 \\ 0 & T \end{matrix} ) ( \begin{matrix}0 & u^{⊺} \\ −u & 0 \end{matrix} )
= ( \begin{matrix}0 & u^{⊺} \\ −v̄ & \underline{ω}T \end{matrix} ) − ( \begin{matrix}0 & 0 \\ 0 & D_{x}T \end{matrix} ) − ( \begin{matrix}0 & u^{⊺} \\ −v̄ & 0 \end{matrix} )
= ( \begin{matrix}0 & 0 \\ 0 & \underline{ω}T − D_{x}T \end{matrix} )

Therefore, to prove the statement, we only need to solve \underline{ω}T = D_{x}T, that is, the matrix T gauges \underline{ω} into zero, which is proved in Appendix B.

4 Hereditary operator in Riemannian Geometry

In this section, we present a hereditary operator that arises in a natural way from the geometric arelength-preserving evolution of curves in n-dimensional Riemannian manifold with constant curvature. The operator is the product of a cosymplectic (Hamiltonian) operator and a symplectic operator.

Theorem 2. Let γ(x, t) be a family of curves on M satisfying a geometric evolution system of equations of the form

γ_{t} = ∑_{l=1}^{n} h^{(l)}e_{l}  (10)

where {e_{l}, l = 1, ⋯, n} is the natural frame of γ, and where h_{l} are arbitrary smooth functions of the curvatures u^{(i)}, i = 1, ⋯ n−1 and their derivatives with respect to x.

Assume that x is the arc-length parameter and that evolution (10) is arclength preserving.

Then, the curvatures u = (u^{(1)}, ⋯, u^{(n−1)})^{⊺} satisfy the evolution

u_{t} = ℜh − κh, h = (h^{(2)}, ⋯, h^{(n)})^{⊺},

where the operator ℜ is hereditary and defined as follows:

ℜ = D_{x}^{2} + ⟨u, u⟩ + u_{1}D_{x}^{−1}⟨u, ·⟩ − ∑_{i<j} J_{ij}uD_{x}^{−1}⟨J_{ij}u_{1}, ·⟩,

and where the J_{ij} are anti-symmetric matrices with nonzero entry of (i, j) being 1 if i < j, that is, (J_{ij})_{kl} = δ_{k}^{i}δ_{j}^{l} − δ_{l}^{i}δ_{k}^{j}. Moreover, ℜ can be written as ℌℑ, where ℑ is a symplectic operator defined by ℑ = D_{x} + uD_{x}^{−1}u^{⊺}, and ℌ is a cosymplectic (or Hamiltonian) operator, defined by

ℌ = D_{x} + ∑_{i<j} J_{ij}uD_{x}^{−1}(J_{ij}u)^{⊺}.

Proof. In the Riemannian case, 𝔥 is 𝔬_{n} and 𝔤 is 𝔢𝔲𝔠_{n}. By fixing the frame, we know the value of ω(D_{x}) and τ(D_{x}). We assumed our frame to be the natural (or parallel) frame, see [Bis75]. So we have e_{1} = X. Due to de_{0} = τ ⊗ e, we know

τ(D_{x}) = (1, 0, ⋯, 0).

The Cartan matrix of the natural frame is by definition

ω(D_{x}) = ( \begin{matrix}0 & u^{(1)} & u^{(2)} & ⋯ & u^{(n−1)} \\ −u^{(1)} & 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ −u^{(n−1)} & 0 & 0 & ⋯ & 0 \end{matrix} ) ,

that is, ω_{ij}(D_{x}) = δ_{1}^{i}u^{(j−1)} − δ_{1}^{j}u^{(i−1)}.

Now we use the moving frame method in section 2 to derive the evolution equation of the curvatures of a smooth curve γ(x, t) in n-dimensional Riemannian manifold satisfying a geometric evolution of the form

γ_{t} = ∑_{i=1}^{n} h^{(i)}e_{i},  (11)

where {e_{i}, i = 1 ⋯ n} is the natural frame and h^{(i)} are arbitrary smooth functions of the curvatures u^{(i)}, i = 1 ⋯ n − 1, and their derivatives with respect to the arclength parameter x.

The given curve (11) implies h^{(i)} = τ_{i}(D_{t}). So the equation (4) reads

T_{i}(D_{x}, D_{t}) = D_{x}h^{(i)} − ω_{1i}(D_{t}) + u^{(i−1)}h^{(1)} − δ_{1}^{i} ∑_{j=2}^{n} h^{(j)}u^{(j−1)}

and we find

ω_{1i}(D_{t}) = D_{x}h^{(i)} + u^{(i−1)}h^{(1)} − δ_{1}^{i} ∑_{j=2}^{n} h^{(j)}u^{(j−1)} − ∑_{j=2}^{n} T_{i}(D_{x}, e_{j})h^{(j)}. (12)

We redefine ω by

ω̃_{1i} = ω_{1i} + T_{i}(D_{x}, ·).

This does not influence our previous calculations since ω(D_{x}) = ω̃(D_{x}), and reflects the fact that we can define the torsionless connection, i.e., Riemannian connection. It is customary to call the geometry Riemannian if the torsion is zero. We now write ω_{1i} for ω̃_{1i} to avoid the complication of the notation and rewrite (12) as

ω_{1i}(D_{t}) = D_{x}h^{(i)} + u^{(i−1)}h^{(1)} − δ_{1}^{i} ∑_{j=2}^{n} h^{(j)}u^{(j−1)}.  (13)

We need ω_{11} = 0 for the consistence, i.e.,

0 = D_{x}h^{(1)} − ∑_{j=2}^{n} h^{(j)}u^{(j−1)}.

Geometrically it means that the evolution is arc-length preserving. By eliminating h^{(1)} in (13), we obtain, taking the integration constants equal to zero,

ω_{1i}(D_{t}) = D_{x}h^{(i)} + u^{(i−1)}D_{x}^{−1} ∑_{j=2}^{n} h^{(j)}u^{(j−1)} = ℑh, 1 < i ≤ n,

with h = (h_{2}, ⋯, h_{n}). This defines a symplectic operator ℑ = D_{x} + uD_{x}^{−1}u^{⊺}, cf. [Wan03], as will be proved in Proposition 1 in Appendix A. We now do the same for 𝔬_{n}-component in (4). Assuming j > i (ω is antisymmetric), we have

Ω_{ij}(D_{x}, D_{t}) = D_{x}ω_{ij}(D_{t}) − D_{t}(δ_{1}^{i}u^{(j−1)} − δ_{1}^{j}u^{(i−1)})

− ∑_{l=1}^{n}(δ_{1}^{i}u^{(l−1)} − δ_{1}^{l}u^{(i−1)})ω_{lj}(D_{t}) + ∑_{l=1}^{n} ω_{il}(D_{t})(δ_{1}^{l}u^{(j−1)} − δ_{1}^{j}u^{(l−1)}).

This leads to

D_{t}u^{(j−1)} = D_{x}ω_{1j}(D_{t}) − ∑_{l=2}^{n} u^{(l−1)}ω_{lj}(D_{t}) − Ω_{1j}(D_{x}, D_{t}), j = 2, ⋯, n

when i = 1, and when i > 1,

D_{x}ω_{ij}(D_{t}) = Ω_{ij}(D_{x}, D_{t}) − u^{(i−1)}ω_{1j}(D_{t}) + u^{(j−1)}ω_{1i}(D_{t}).

We combine these to

D_{t}u^{(j−1)} = −Ω_{1j}(D_{x}, D_{t}) + D_{x}ω_{1j}(D_{t})

− ∑_{l=2}^{n} u^{(l−1)}D_{x}^{−1}(−u^{(l−1)}ω_{1j}(D_{t}) + u^{(j−1)}ω_{1l}(D_{t})) j = 2, ⋯, n

Assume that the curvature of Riemannian manifold is constant, i.e., the only nonzero entries in Ω(e_{i}, e_{j}) are

Ω_{ij}(e_{i}, e_{j}) = −Ω_{ji}(e_{i}, e_{j}) = κ, j > i.

Then Ω_{1j}(D_{x}, D_{t}) = h_{j}. The above formula defines in the case of constant curvature a cosymplectic (or Hamiltonian) operator ℌ (cf. [Wan03]) given by

ℌ = D_{x} + ∑_{i<j} J_{ij}uD_{x}^{−1}(J_{ij}u)^{⊺},

where the J_{ij}, given by (J_{ij})_{kl} = δ_{ik}δ_{jl} − δ_{il}δ_{jk}, are the generators of 𝔬_{n} and such that D_{t}u = ℌω_{1·}(D_{t}). The fact that the operator is indeed Hamiltonian will be given in Proposition 2 in Appendix A. The operator is weakly non-local, as defined in [MN01].

It corresponds to what is called the second Poisson operator P_{u}, defined by

P_{u} = D_{x} + π_{1}ad(u) + ad(u)D_{x}^{−1}π_{0}ad(u),

in [TT01], but looks simpler by the choice of frame.

We write the evolution equation for the curvatures u^{(j)} as

u_{t} = ℜh − κh,

where the operator ℜ = ℌℑ is explicitly given as

ℜ = D_{x}^{2} + ⟨u, u⟩ + u_{1}D_{x}^{−1}⟨u, ·⟩ − ∑_{i<j} J_{ij}uD_{x}^{−1}⟨J_{ij}u_{1}, ·⟩.  (14)

We prove that ℜ is hereditary in Proposition 3 in Appendix A.

Remark 1. The recursion operator can be used to generate infinitely many compatible symplectic and cosymplectic weakly non-local operators starting with ℑ and ℌ. That the result is indeed weakly non-local follows from the techniques in [SW01a, MN01].

Remark 2. A special case arises for n = 3: one has

ℜ = −(J_{12}(D_{x} + uD_{x}^{−1}u))^{2} = −R_{NLS}^{2},

where R_{NLS} is the recursion operator of the Nonlinear Schrödinger (NLS) equation.

Let us now take h^{(j)} = u_{1}^{(j−1)} in Theorem 2. Then we obtain

u_{t} = u_{3} + \frac{3}{2}⟨u, u⟩u_{1} − κu_{1} (vmKDV),

which is one of two versions of vector mKDV equations appearing in [SW01b].

Corollary 1. The hereditary operator defined in (14) is a recursion operator of the vector mKDV equation u_{t} = u_{3} + \frac{3}{2}⟨u, u⟩u_{1}.

Although we give the explicit formula of the generalized Hasimoto transformation in section 3, it is difficult to use it to compute the formula (1). It is easier to obtain it directly using the moving frame method in section 2.

Theorem 3. Let us assume that in the Frenêt frame, the curvatures of a geometric curve γ_{t} = ∑_{l=1}^{n} h̄^{(l)}ē_{l} under arc-length preserving satisfy

ū_{t} = Φ̄_{1}h̄ − κΦ̄_{2}h̄, h̄ = (h̄^{(2)}, ⋯, h̄^{(n)})^{⊺}.  (15)

There exists an operator Q such that Φ̄_{1}Q and Φ̄_{2}Q are compatible Hamiltonian operators and Q = T ℌT^{⊺}Φ̄_{2}^{∗}, where ℌ is defined in Theorem 2 and T is defined in the proof of Theorem 1.

Proof. First we notice that h̄ = T h in Riemannian manifold. The Hasimoto transformation is also a Miura transformation between u_{t} and ū_{t}, that is,

u_{t} = D_{u} ū_{t} = (ℜ − κ)h = D_{u}(Φ̄_{1} − κΦ̄_{2})h̄.

Therefore, D_{u}^{−1} = Φ̄_{2}T and ℜ = D_{u} Φ̄_{1}T. We know that ℌ and ℜℌ are compatible Hamiltonian operators. Thus

D_{u}^{−1}ℌD_{u}^{−1,∗} = Φ̄_{2}T ℌT^{⊺}Φ̄_{2}^{∗} and D_{u}^{−1}ℜℌD_{u}^{−1,∗} = Φ̄_{1}T ℌT^{⊺}Φ̄_{2}^{∗}

are also compatible. By now, we prove the statement.

The operator Q can be computed explicitly and it is independent on the angles in the Hasimoto transformation.

5 A Lax pair of vector mKDV

For the Lie algebra background to find Lax pairs, see [FK83, For90]. The computation of the Lax pair is known in the literature. In the case of 𝔬_{n}, two derivations, both using different ℤ/2-gradings, are known to us: [TT01, LP00]. Both derivations at some point use some not so obvious steps and we noted that the steps that were obvious in one were not obvious in the other. So it seemed like a natural idea to use both ℤ/2-gradings and do the derivation in a completely obvious way.

We now identify e_{1} with an element in 𝔬_{n+1} which we call L^{10} using the identification

𝔢𝔲𝔠_{n} ϶ ( \begin{matrix}0 & v^{⊺} \\ 0 & ω \end{matrix} ) ↦^{ϕ} ( \begin{matrix}0 & v^{⊺} \\ −v & ω \end{matrix} ) ∈ 𝔬_{n+1}

Let L = ϕ_{⋆}A(D_{x}), where A is defined in equation (3) and let λ be the norm of the tangent vector to the curve, that is, this vector equals λe_{1}. Then

L = L^{01} + λL^{10} =

= ( \begin{matrix}0 & 0, & 0, & ⋯, & 0 \\ 0 & 0, & u^{(1)}, & ⋯, & u^{(n−1)} \\ 0 & −u^{(1)}, & 0, & ⋯, & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ 0 & −u^{(n−1)}, & 0, & ⋯, & 0 \end{matrix} ) + λ ( \begin{matrix}0 & 1, & 0, & ⋯, & 0 \\ −1 & 0, & 0, & ⋯, & 0 \\ 0 & 0, & 0, & ⋯, & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ 0 & 0, & 0, & ⋯, & 0 \end{matrix} ) .

In order to get some control over our calculations, we introduce two ℤ/2gradings. We do this by partitioning the n + 1-dimensional matrix into (1, n) × (1, n)-blocks, and into (2, n − 1) × (2, n − 1)-blocks. We then grade the diagonal blocks with 0 and the off-diagonal blocks with 1. This way we can write 𝔬_{n+1} as

𝔬_{n+1} = 𝔤^{00} + 𝔤^{01} + 𝔤^{10} + 𝔤^{11}.

We see that L^{01} ∈ 𝔤^{01} and L^{10} ∈ 𝔤^{10}, which explains the notation. Observe that the first grading is consistent with the parity of the power of λ, the second is not. So one does expect the computation to be homogeneous in the first grading and inhomogeneous with respect to the second. The second grading codes some useful facts. If X ∈ ker ad(L^{10}), then X ∈ 𝔤^{·0} and if X ∈ im ad(L^{10}), then X ∈ 𝔤^{·1}. If X ∈ 𝔤^{10}, then X ∈ ⟨L^{10}⟩_{ℝ}. Furthermore, ad(L^{10}) gives us a complex structure on 𝔤^{01} + 𝔤^{10}. We now let A(D_{t}) = M = M_{3} + λM_{2} + λ^{2}M_{1} + λ^{3}M_{0}. The reader may want to experiment with lower order expansions of M.

From the curvature equation (5)

F (D_{x}, D_{t}) = [D_{x} + L, D_{t} + M]

it follows that the λ^{4} term vanishes, that is: ad(L^{10})M_{0} = 0. We simply choose M_{0} = −L^{10}. The next order λ^{3} gives us

ad(L^{10})M_{1} = −D_{x}M_{0} − [L^{01}, M_{0}].

or M_{1} + L^{01} ∈ ker ad(L^{10}). We take M_{1} = −L^{01}.

On the λ^{2} level we find

ad(L^{10})M_{2} = −D_{x}M_{1} − [L^{01}, M_{1}] = −D_{x}L^{01}.

We can solve this by letting M_{2} = −D_{x}ad(L^{10})L^{01}+βL^{10}, with β some arbitrary function. We find on the λ level

ad(L^{10})M_{3} = D_{x}^{2}ad(L^{10})L^{01} − β_{x}L^{10} + [L^{01}, D_{x}ad(L^{10})L^{01}] − β[L^{01}, L^{10}]

or

ad(L^{10})M_{3} = ad(L^{10})(D_{x}^{2}L^{01} + βL^{01}) − β_{x}L^{10} − ad(L^{01})ad(D_{x}L^{01})L^{10}

Looking at the grading, we see that ad(L^{01})ad(D_{x}L^{01})L^{10} = κ(L^{01}, D_{x}L^{01})L^{10}, where κ is a bilinear form with values in ℝ. We find that on the span of L^{10} and L^{01} the operator ad(L^{01})ad(D_{x}L^{01}) behaves like

⟨u, u⟩D_{x} − ⟨u, u_{1}⟩.

In other words,

ad(L^{10})M_{3} = ad(L^{10})(D_{x}^{2}L^{01} + βL^{01}) − β_{x}L^{10} + ⟨u, u_{1}⟩L^{10}

So we take β = ν + \frac{1}{2}⟨u, u⟩, ν ∈ ℝ. It follows that M_{3} = D_{x}^{2}L^{01} + (ν + \frac{1}{2}⟨u, u⟩)L^{01} + X^{00}. The last equation (and this gives us the obstruction to solving the structure equations) reads

F (D_{x}, D_{t}) = D_{x}M_{3} − D_{t}L^{01} + [L^{01}, M_{3}].

This implies

D_{t}L^{01} = D_{x}^{3}L^{01} + ⟨u, u_{1}⟩L^{01} + (ν + \frac{1}{2}⟨u, u⟩)D_{x}L^{01} + D_{x}X^{00}
+D_{x}[L^{01}, D_{x}L^{01}] + [L^{01}, X^{00}] − F (D_{x}, D_{t})

We take X^{00} = −[L^{01}, D_{x}L^{01}] and obtain

D_{t}L^{01} = D_{x}^{3}L^{01} + ⟨u, u_{1}⟩L^{01} + (ν + \frac{1}{2}⟨u, u⟩)D_{x}L^{01}
+ad(L^{01})ad(D_{x}L^{01})L^{01} − F (D_{x}, D_{t})
= D_{x}^{3}L^{01} + ⟨u, u_{1}⟩L^{01} + (ν + \frac{1}{2}⟨u, u⟩)D_{x}L^{01}
+(⟨u, u⟩D_{x} − ⟨u, u_{1}⟩)L^{01} − F (D_{x}, D_{t})
= D_{x}^{3}L^{01} + (ν + \frac{3}{2}⟨u, u⟩)D_{x}L^{01} − F (D_{x}, D_{t})

Further computation shows that the Killing form of L^{01} is given by

K(L^{01}, L^{01}) = −2(n − 2)⟨u, u⟩,

and we can write, at least for n > 2,

D_{t}L^{01} = D_{x}^{3}L^{01} + (ν − \frac{3}{4(n − 2)}K(L^{01}, L^{01}))D_{x}L^{01} − F (D_{x}, D_{t}).

We now make the choice D_{t} = ∑_{i=1}^{n−1} u^{(i)}u_{1}^{(i)}e_{1} +∑_{i=1}^{n−1} u_{1}^{(i)}e_{i+1}, which is consistent with the condition ω_{11} = 0. We let R_{ijkl} = Ω_{ij}(e_{k}, e_{l}). Constant curvature means that R_{ijij} = κ for i ≠ j and the other R coefficients are zero. We obtain

D_{t}L^{01} = D_{x}^{3}L^{01} + (ν − \frac{3}{4(n − 2)}K(L^{01}, L^{01}))D_{x}L^{01} − κD_{x}L^{01}.

Observe that the role of the integration constant and the curvature is the same. This may explain some of the success of the zero-curvature method.

We can flatten the equation by absorbing the curvature κ in the connection by taking the integration constant ν equal to κ:

D_{t}L^{01} = D_{x}^{3}L^{01} − \frac{3}{4(n − 2)}K(L^{01}, L^{01})D_{x}L^{01}

or

u_{t} = u_{3} + \frac{3}{2}||u||^{2}u_{1} (vmKDV).

From the point of view of integrability this flattening does not have any influence, since u_{1} is a trivial symmetry of the equation. We remark that while we have found the vmKDV equation through the Lax pair construction, it also provides a Lax pair for vmKDV, taking its values in 𝔬_{m+1}, if u is an (m−1)-vector, cf. [TT01].

A Geometric structures

In this Appendix we collect the proofs that the operators ℑ, ℌ and ℜ are symplectic, cosymplectic and hereditary, respectively. The fact that ℜ is hereditary is equivalent to the compatibility of the Hamiltonian structures ℑ^{−1} and ℌ.

These operators have nonlocal terms by the occurrence of D_{x}^{−1}. We refer to [SW01a] for results to on the locality of the generated symmetries and cosymmetries.

Proposition 1. The operator ℑ = D_{x} + uD_{x}^{−1}u^{⊺} is symplectic.

Proof. As is proved in [Dor93], to show that ℑ is symplectic it suffices that

∫ (⟨ℑ′[h_{1}]h_{2}, h_{3}⟩ + ⟨ℑ′[h_{3}]h_{1}, h_{2}⟩ + ⟨ℑ′[h_{2}]h_{3}, h_{1}⟩)dx = 0.

One has ℑ′[h_{1}] = h_{1}D_{x}^{−1}u^{⊺} + uD_{x}^{−1}h_{1}^{⊺}. Then

⟨ℑ′[h_{1}]h_{2}, h_{3}⟩ =
= ⟨h_{1}D_{x}^{−1}u^{⊺}h_{2}, h_{3}⟩ + ⟨uD_{x}^{−1}h_{1}^{⊺}h_{2}, h_{3}⟩
= D_{x}^{−1}(⟨u, h_{2}⟩)⟨h_{1}, h_{3}⟩ + D_{x}^{−1}(⟨h_{1}, h_{2}⟩)⟨u, h_{3}⟩
≡ D_{x}^{−1}(⟨u, h_{2}⟩)⟨h_{1}, h_{3}⟩ − D_{x}^{−1}(⟨u, h_{3}⟩)⟨h_{1}, h_{2}⟩,

and the result follows upon summation of the 3 cyclically permuted terms.

Proposition 2. The operator ℌ, given by

ℌ = D_{x} + ∑_{i<j} J_{ij}uD_{x}^{−1}(J_{ij}u)^{⊺},

where the J_{ij}, given by (J_{ij})_{kl} = δ_{ik}δ_{jl} − δ_{il}δ_{jk}, are the generators of 𝔬_{n−1}, is a cosymplectic (or Hamiltonian) operator.

Proof. We prove this by checking the conditions of Theorem 7.8 and Corollary 7.21 in [Olv93].

The associated bi-vector of ℌ is by definition

Θ_{ℌ} = \frac{1}{2} ∫ (θ ˙∧ℌθ) dx

= \frac{1}{2} ∫ (θ ˙∧ (θ_{1} + ∑_{i<j} J_{ij}uD_{x}^{−1}(J_{ij}u)^{⊺}θ)) dx

= \frac{1}{2} ∫ θ ˙∧θ_{1}dx + \frac{1}{2} ∑_{i<j} ∫ ⟨J_{ij}u, θ⟩ ∧ ⟨J_{ij}u, θ⟩_{−1}dx,

where θ = (ϑ^{1}, ⋯, ϑ^{n}) and ϑ_{i}^{j} = \frac{∂^{i}ϑ^{j}}{∂x^{i}}, etc. Here ˙∧ means that one needs to take the ordinary inner product between the vectors θ and ℌθ. The elements of these vectors are then multiplied using the ordinary wedge product. We need to check the vanishing of the Schouten bracket [ℌ, ℌ] which is equivalent to the Jacobi identity for the Lie bracket defined by ℌ.

[ℌ, ℌ] = Pr v_{ℌθ}(Θ_{ℌ}) =

= ∑_{i<j} ∫ (⟨J_{ij}ℌθ, θ⟩ ∧ ⟨J_{ij}u, θ⟩_{−1}) dx

= ∑_{i<j} ∫ J_{ij}θ_{1} ˙∧θ ∧ ⟨J_{ij}u, θ⟩_{−1}dx

+ ∑_{i<j} ∑_{k<l} ∫ ⟨J_{kl}u, θ⟩_{−1}J_{ij}J_{kl}u ˙∧θ ∧ ⟨J_{ij}u, θ⟩_{−1}dx.

The first term is zero because

J_{ij}θ_{1} ˙∧θ ∧ ⟨J_{ij}u, θ⟩_{−1} =
= (θ_{1}^{j} ∧ θ^{i} − θ_{1}^{i} ∧ θ^{j}) ∧ D_{x}^{−1}(u^{j}θ^{i} − u^{i}θ^{j})
= (θ_{1}^{j} ∧ θ^{i} + θ^{j} ∧ θ_{1}^{i}) ∧ D_{x}^{−1}(u^{j}θ^{i} − u^{i}θ^{j})
≡ −θ^{j} ∧ θ^{i} ∧ (u^{j}θ^{i} − u^{i}θ^{j})
= 0.

The second term is zero because

∑_{i<j} ∑_{k<l}⟨J_{kl}u, θ⟩_{−1} ∧ ⟨J_{ij}J_{kl}u, θ⟩ ∧ ⟨J_{ij}u, θ⟩_{−1}

= \frac{1}{2} ∑_{i<j} ∑_{k<l}⟨J_{kl}u, θ⟩_{−1} ∧ ⟨[J_{ij}, J_{kl}]u, θ⟩ ∧ ⟨J_{ij}u, θ⟩_{−1}
= ∑_{i<j,k<l,i<k}⟨J_{kl}u, θ⟩_{−1} ∧ ⟨δ_{lj}J_{ik}u, θ⟩ ∧ ⟨J_{ij}u, θ⟩_{−1}
− ∑_{i<j,k<l,i<l}⟨J_{kl}u, θ⟩_{−1} ∧ ⟨δ_{jk}J_{il}u, θ⟩ ∧ ⟨J_{ij}u, θ⟩_{−1}
+ ∑_{i<j,k<l,j<l}⟨J_{kl}u, θ⟩_{−1} ∧ ⟨δ_{ik}J_{jl}u, θ⟩ ∧ ⟨J_{ij}u, θ⟩_{−1}
= ∑_{i<j<k} D_{x} (⟨J_{jk}u, θ⟩_{−1} ∧ ⟨J_{ij}u, θ⟩_{−1} ∧ ⟨J_{ik}u, θ⟩_{−1})
≡ 0.

Thus the result follows. Proposition 3. The operator

ℜ = D_{x}^{2} + ⟨u, u⟩ + u_{1}D_{x}^{−1}⟨u, ·⟩ − ∑_{i<j} J_{ij}uD_{x}^{−1}⟨J_{ij}u_{1}, ·⟩

is hereditary. Proof. Notice that for the matrices J_{ij} we have

∑_{i<j}^{n} J_{ij}uD_{x}^{−1}⟨J_{ij}u_{1}, P ⟩ = (D_{x}^{−1}P u_{1}^{⊺})u − (D_{x}^{−1}u_{1}P^{⊺})u.  (16)

We recall that an hereditary operator ℜ is characterized by the property that ℜD_{ℜ}[P](Q) − D_{ℜ}[ℜP](Q) is symmetric with respect to vectors P and Q, cf. [FF82]. We first compute D_{ℜ}[P](Q), meanwhile we drop the terms that are symmetric with respect to P and Q by using ”≡”.

D_{ℜ}[P](Q) = 2⟨u, P ⟩Q + P_{x}D_{x}^{−1}⟨u, Q⟩ + u_{1}D_{x}^{−1}⟨P, Q⟩

− ∑_{i<j}^{n} J_{ij}P D_{x}^{−1}⟨J_{ij}u_{1}, Q⟩ − ∑_{i<j}^{n} J_{ij}uD_{x}^{−1}⟨J_{ij}P_{x}, Q⟩

≡ ⟨u, P ⟩Q + P_{x}D_{x}^{−1}⟨u, Q⟩ − ∑_{i<j}^{n} J_{ij}P D_{x}^{−1}⟨J_{ij}u_{1}, Q⟩,

where we use the relation ∑_{i<j} J_{ij}uD_{x}^{−1}⟨J_{ij}P_{x}, Q⟩ ≡ ⟨u, P ⟩Q, which can be proved by applying formula (16). Notice that

ℜP = P_{xx} + ⟨u, u⟩P + u_{1}D_{x}^{−1}⟨u, P ⟩ − ∑_{i<j} J_{ij}uD_{x}^{−1}⟨J_{ij}u_{1}, P ⟩

and the terms in ℜD_{ℜ}[P](Q) − D_{ℜ}[ℜP](Q) are u–degree of 1 or 3. Its first degree terms are

⟨u_{2}, P ⟩Q + 2⟨u_{1}, P_{x}⟩Q + ⟨u, P_{xx}⟩Q +2⟨u_{1}, P ⟩Q_{x} + 2⟨u, P_{x}⟩Q_{x} + ⟨u, P ⟩Q_{xx} +P_{x}⟨u, Q_{x}⟩ + P_{x}⟨u_{1}, Q⟩ + 2P_{xx}⟨u, Q⟩ + P_{xxx}D_{x}^{−1}⟨u, Q⟩

− ∑_{i<j} J_{ij}P ⟨J_{ij}u_{2}, Q⟩ − ∑_{i<j} J_{ij}P ⟨J_{ij}u_{1}, Q_{x}⟩

−2 ∑_{i<j} J_{ij}P_{x}⟨J_{ij}u_{1}, Q⟩ − ∑_{i<j} J_{ij}P_{xx}D_{x}^{−1}⟨J_{ij}u_{1}, Q⟩

−2⟨P_{xx}, u⟩Q − P_{xxx}D_{x}^{−1}⟨u, Q⟩ − u_{1}D_{x}^{−1}⟨P_{xx}, Q⟩

+ ∑_{i<j} J_{ij}P_{xx}D_{x}^{−1}⟨J_{ij}u_{1}, Q⟩ + ∑_{i<j} J_{ij}uD_{x}^{−1}⟨J_{ij}P_{xxx}, Q⟩

≡ 0.

Similarly, we can check its third degree terms are also equivalent to zero. Therefore,

(ℜD_{ℜ}[P](Q))^{j} − (D_{ℜ}[ℜP](Q))^{j} ≡ 0,

that is, the operator ℜ is hereditary.

B Generalized Hasimoto transformation

In this Appendix we complete the proof of Theorem 1. The formula contain possibly non-existing angles. If this is the case the convention is to take the first term in the product to be equal to one.

Lemma 1. The matrix T gauges \underline{ω} into zero, cf. the proof of Theorem 1 for the notations of T and \underline{ω}.,that is \underline{ω}T = D_{x}T.

Proof. Notice that R = R_{n−1,n} ⋯ R_{2n}R_{n−2,n−1} ⋯ R_{34}R_{24}R_{23} since R_{ij} commutes R_{kl} when {i, j} ⋂{k, l} = ∅. Let R^{(i)} = R_{i−1,i} ⋯ R_{3i}R_{2i} and the part without the first row and the first column is denoted by T^{(i)}. Then one writes R = R^{(n)}R^{(n−1)} ⋯ R^{(4)}R^{(3)} and T = T^{(n)}T^{(n−1)} ⋯ T^{(4)}T^{(3)}. The matrix T^{(i)} is generated by

T_{jj}^{(i)} = 1, j > i − 1
T_{jj}^{(i)} = cos θ_{j+1,i}, 1 < j < i − 1,
T_{1,i−1}^{(i)} = sin θ_{2i}
T_{lj}^{(i)} = sin θ_{l+1,i} \frac{∂T_{l−1,j}^{(i)}}{∂θ_{l,i}} , j < l < i − 1
T_{i−1,j}^{(i)} = \frac{∂T_{i−2,j}^{(i)}}{∂θ_{i−1,i}} .

and the rest entries are equal to zero.

We first perform the gauge transformation of T^{(n)} on \underline{ω}. Notice the last column of \underline{ω}T^{(n)} − D_{x}T^{(n)} has to be zero since the others do not affect it. This implies that

ū^{(2)}T_{2,n−1}^{(n)} − D_{x}T_{1,n−1}^{(n)} = 0;  (17)
ū^{(k)}T_{k,n−1}^{(n)} − ū^{(k−1)}T_{k−2,n−1}^{(n)} − D_{x}T_{k−1,n−1}^{(n)} = 0, 2 < k ≤ n − 1; (18)
ū^{(n−1)}T_{n−2,n−1}^{(n)} + D_{x}T_{n−1,n−1}^{(n)} = 0,  (19)

where T_{j,n−1}^{(n)} = sin θ_{j+1,n} cos θ_{j,n} ⋯ cos θ_{2n} and 1 ≤ j ≤ n − 1. The formula (17) immediately leads to

ū^{(2)} = \frac{D_{x}θ_{2n}}{sin θ_{3n}} .

Now we prove the following formula by induction on i.

ū^{(i)} = \frac{D_{x}θ_{i,n}}{sin θ_{i+1,n}} + \frac{cos θ_{i,n}}{sin θ_{i+1,n}} \frac{sin θ_{i−1,n}}{cos θ_{i−1,n}} ū^{(i−1)}, 2 < i ≤ n − 1. (20)

Using (18) and assuming that (20) is true for 2 < l ≤ k − 1, we compute

ū^{(k)}T_{k,n−1}^{(n)} = ū^{(k)} sin θ_{k+1,n} cos θ_{k,n} ⋯ cos θ_{2,n}

= ū^{(k−1)}T_{k−2,n−1}^{(n)} + D_{x}T_{k−1,n−1}^{(n)}
= ū^{(k−1)} sin θ_{k−1,n} cos θ_{k−2,n} ⋯ cos θ_{2,n}

+ cos θ_{k,n} cos θ_{k−1,n} ⋯ cos θ_{2,n}D_{x}θ_{k,n}

− ∑_{l=2}^{k−1} sin θ_{k,n} cos θ_{k−1,n} ⋯ sin θ_{l,n} ⋯ cos θ_{2,n}D_{x}θ_{l,n}

= ū^{(k−1)} cos^{2} θ_{k,n} sin θ_{k−1,n} cos θ_{k−2,n} ⋯ cos θ_{2,n}

+ cos θ_{k,n} cos θ_{k−1,n} ⋯ cos θ_{2,n}D_{x}θ_{k,n}.

Thus, we proved the formula is valid for ū^{(k)}. By directly computation, we can check the identity (19).

Next, we prove \underline{ω}T^{(n)} − D_{x}T^{(n)} = T^{(n)}C, where C is an anti-symmetric matrix with

C_{k−1,k} = −C_{k,k−1} = ∑_{l=2}^{k} \frac{cos θ_{k+1,n}}{sin θ_{k+1,n}} \frac{1}{sin θ_{k,n}} \frac{sin θ_{l,n}}{cos θ_{l,n}} D_{x}θ_{l,n}, 2 ≤ i ≤ n − 2,

and the remaining entries are equal to zero.

To do so, we first rewrite (20) as

ū^{(i)} = ∑_{l=2}^{i} \frac{1}{sin θ_{i+1,n}} \frac{cos θ_{i,n}}{sin θ_{i,n}} \frac{sin θ_{l,n}}{cos θ_{l,n}} D_{x}θ_{l,n}, 2 ≤ i ≤ n − 1. (21)

Let us check that

0 = ū^{(i+1)}T_{i+1,j}^{(n)} − ū^{(i)}T_{i−1,j}^{(n)} − D_{x}T_{i,j}^{(n)}
− T_{i,j−1}^{(n)}C_{j−1,j} + T_{i,j+1}^{(n)}C_{j,j+1}.  (22)

It is easy to see that this formula is valid for j > i+1 since T_{l,k}^{(n)} = 0 when l < k and k ≠ n − 1. When j = i + 1, we have

ū^{(i+1)}T_{i+1,i+1}^{(n)} − T_{i,i}^{(n)}C_{i,i+1} = ū^{(i+1)} cos θ_{i+2,n} − cos θ_{i+1,n}C_{i,i+1} = 0.

Similarly, for j = i, we can check ū^{(i+1)}T_{i+1,i}^{(n)} − D_{x}T_{i,i}^{(n)} − T_{i,i−1}^{(n)}C_{i−1,i} = 0. Now we concentrate on the case j < i. Notice that

T_{i,j}^{(n)} = − sin θ_{i+1,n} cos θ_{i,n} ⋯ cos θ_{j+2,n} sin θ_{j+1,n}, i > j.

So, we have

ū^{(i+1)}T_{i+1,j}^{(n)} − ū^{(i)}T_{i−1,j}^{(n)} − D_{x}T_{i,j}^{(n)} − T_{i,j−1}^{(n)}C_{j−1,j} + T_{i,j+1}^{(n)}C_{j,j+1}

= ∑_{l=2}^{i+1} \frac{1}{sin θ_{i+2,n}} \frac{cos θ_{i+1,n}}{sin θ_{i+1,n}} \frac{sin θ_{l,n}}{cos θ_{l,n}} D_{x}θ_{l,n}(− sin θ_{i+2,n} ∏_{k=j+2}^{i+1} cos θ_{k,n} sin θ_{j+1,n})

− ∑_{l=2}^{i} \frac{1}{sin θ_{i+1,n}} \frac{cos θ_{i,n}}{sin θ_{i,n}} \frac{sin θ_{l,n}}{cos θ_{l,n}} D_{x}θ_{l,n}(− sin θ_{i,n} ∏_{k=j+2}^{i−1} cos θ_{k,n} sin θ_{j+1,n})

−D_{x} (− sin θ_{i+1,n} ∏_{k=j+2}^{i} cos θ_{k,n} sin θ_{j+1,n})

−(− sin θ_{i+1,n} ∏_{k=j+1}^{i} cos θ_{k,n} sin θ_{j,n}) ∑_{l=2}^{j} \frac{cos θ_{j+1,n}}{sin θ_{j+1,n}} \frac{1}{sin θ_{j,n}} \frac{sin θ_{l,n}}{cos θ_{l,n}} D_{x}θ_{l,n}

+(− sin θ_{i+1,n} ∏_{k=j+3}^{i} cos θ_{k,n} sin θ_{j+2,n}) ∑_{l=2}^{j+1} \frac{cos θ_{j+2,n}}{sin θ_{j+2,n}} \frac{1}{sin θ_{j+1,n}} \frac{sin θ_{l,n}}{cos θ_{l,n}} D_{x}θ_{l,n}

= 0.

Now we apply induction procedure on the matrix C since it is of the form of Frenêt frame. Using (21) for the n − 1 case, we have

C_{k−1,k} = ∑_{l=2}^{k} \frac{1}{sin θ_{k+1,n−1}} \frac{cos θ_{k,n−1}}{sin θ_{k,−1n}} \frac{sin θ_{l,n−1}}{cos θ_{l,n−1}} D_{x}θ_{l,n−1}, 2 ≤ k ≤ n − 2.

Form these, we can solve D_{x}θ_{k,n} for 2 ≤ k ≤ n − 2.

D_{x}θ_{k,n} = ∑_{l=2}^{k} \frac{sin θ_{k+1,n}}{cos θ_{k+1,n}} \frac{cos θ_{k,n}}{sin θ_{k+1,n−1}} \frac{cos θ_{k,n−1}}{sin θ_{k,n−1}} \frac{sin θ_{l,n−1}}{cos θ_{l,n−1}} D_{x}θ_{l,n−1}

− ∑_{l=2}^{k−1} \frac{cos θ_{k,n}}{sin θ_{k,n}} \frac{sin θ_{l,n}}{cos θ_{l,n}} D_{x}θ_{l,n} 2 ≤ k ≤ n − 2.

Assume that the formula (9) is valid for j = n − 1, and 2 ≤ i ≤ n − 3. When k = 2, we obtain

D_{x}θ_{2,n} = \frac{sin θ_{3,n}}{cos θ_{3,n}} \frac{cos θ_{2,n}}{sin θ_{3,n−1}} ∏_{l=4}^{n−1} \frac{cos θ_{2,l}}{cos θ_{3,l}} sin θ_{3,n−1}A_{2}

= ∏_{l=4}^{n} \frac{cos θ_{2,l}}{cos θ_{3,l}} sin θ_{3,n}A_{2}.

Now assuming that the formula (9) is valid for j = n, and 2 ≤ i ≤ k − 1, we compute

D_{x}θ_{k,n} = ∑_{l=2}^{k} \frac{sin θ_{k+1,n}}{cos θ_{k+1,n}} \frac{cos θ_{k,n}}{sin θ_{k+1,n−1}} \frac{cos θ_{k,n−1}}{sin θ_{k,n−1}} \frac{sin θ_{l,n−1}}{cos θ_{l,n−1}} ·

( ∏_{r=l+2}^{n−1} \frac{cos θ_{lr}}{cos θ_{l+1,r}} sin θ_{l+1,n−1}A_{l} − ∏_{r=l+1}^{n−2} \frac{cos θ_{l−1,r}}{cos θ_{l,r}} sin θ_{l−1,n−1}A_{l−1})

− ∑_{l=2}^{k−1} \frac{cos θ_{k,n}}{sin θ_{k,n}} \frac{sin θ_{l,n}}{cos θ_{l,n}} ·

( ∏_{r=l+2}^{n} \frac{cos θ_{lr}}{cos θ_{l+1,r}} sin θ_{l+1,n}A_{l} − ∏_{r=l+1}^{n−1} \frac{cos θ_{l−1,r}}{cos θ_{l,r}} sin θ_{l−1,n}A_{l−1}, )

= ∏_{r=k+2}^{n} \frac{cos θ_{kr}}{cos θ_{k+1,r}} sin θ_{k+1,n}A_{k} − ∏_{r=k+1}^{n−1} \frac{cos θ_{k−1,r}}{cos θ_{k,r}} sin θ_{k−1,n}A_{k−1}.

Therefore, (9) is also valid for j = n and i = k.

Substituting (9) into (21), we have

ū^{(i)} = ∑_{l=2}^{i} \frac{1}{sin θ_{i+1,n}} \frac{cos θ_{i,n}}{sin θ_{i,n}} \frac{sin θ_{l,n}}{cos θ_{l,n}} ·

( ∏_{r=l+2}^{n} \frac{cos θ_{lr}}{cos θ_{l+1,r}} sin θ_{l+1,n}A_{l} − ∏_{r=l+1}^{n−1} \frac{cos θ_{l−1,r}}{cos θ_{l,r}} sin θ_{l−1,n}A_{l−1})

= ∏_{r=i+2}^{n} \frac{cos θ_{ir}}{cos θ_{i+1,r}} A_{i}.

Finally, we prove (8) for i = n − 1. Taking i = n − 1 in (20), we have

ū^{(n−1)} = D_{x}θ_{n−1,n} + cos θ_{n−1,n} \frac{sin θ_{n−2,n}}{cos θ_{n−2,n}} ū^{(n−2)}

On the other hand, we have ū^{(n−1)} = A_{n−1} and ū^{(n−2)} = \frac{cos θ_{n−2,n}}{cos θ_{n−1,n}} A_{n−2}. Therefore

A_{n−1} = D_{x}θ_{n−1,n} + sin θ_{n−2,n}A_{n−2}.

By now we completely proved the statement.

C Integrable systems in 3 dimensional Riemannian geometry

In this appendix we specialize to n = 3. We show that the operator Q in theorem 3 can be explicitly constructed. We make a comparison with known results, [MBSW02] and [DS94]. The equations are (cf. [DS94])

ū_{t} = Ψ̄_{1}h̄ − κΨ̄_{2}h̄,

where, with ū^{⊺} = (κ, τ),

Ψ̄_{1} = ( \begin{matrix}D_{x}^{2} + D_{x}κD_{x}^{−1}κ − τ^{2} & −D_{x}τ − τD_{x} \\ D_{x} \frac{τ}{κ}D_{x} + D_{x}τD_{x}^{−1}κ + D_{x} \frac{1}{κ}D_{x}τ + κτ & D_{x} \frac{1}{κ}D_{x}^{2} − D_{x} \frac{τ^{2}}{κ} + κD_{x} \end{matrix} ) .

and

Ψ̄_{2} = ( \begin{matrix}1 & 0 \\ 0 & D_{x} \frac{1}{κ} \end{matrix} ) .

We know from the Hasimoto transformation (6) that

γ_{t} = (h̄_{1}, h̄_{2}, h̄_{3}) ( \begin{matrix}T \\ N \\ B \end{matrix} ) = (h_{1}, h_{2}, h_{3}) ( \begin{matrix}T \\ N^{1} \\ N^{2} \end{matrix} )

= (h_{1}, h_{2}, h_{3}) ( \begin{matrix}1 & 0 & 0 \\ 0 & cos θ & − sin θ \\ 0 & sin θ & cos θ \end{matrix} ) ( \begin{matrix}T \\ N \\ B \end{matrix} )

or

(h̄_{2}, h̄_{3}) = (h_{2}, h_{3}) ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) = (h_{2}, h_{3},)T

In this case, we know that H = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) is a Hamiltonian operator of (10).

Q = T HT^{⊺}Ψ_{2}^{⋆}
= ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) ( \begin{matrix}1 & 0 \\ 0 & −\frac{1}{κ}D_{x} \end{matrix} )
= − ( \begin{matrix}0 & \frac{1}{κ}D_{x} \\ 1 & 0 \end{matrix} ) .

Using the notation in [MBSW02] we have (Ψ̄_{1} + κΨ̄_{2})Q = −P. This explains the rather mysterious result that P is a compatible Hamiltonian operator.

D Integrable systems in 4 dimensional Rieman-

nian geometry

In this appendix we specialize to n = 4. The equations are

ū_{t} = Ψ̄_{1}h̄ − κΨ̄_{2}h̄,

where, with ū^{⊺} = (u, v, w),

Ψ̄_{2} = ( \begin{matrix}1 & 0 & 0 \\ 0 & D_{x} \frac{1}{u} & −\frac{w}{u} \\ 0 & D_{x} \frac{w}{uv} & D_{x} \frac{1}{v} D_{x} \frac{1}{u} + \frac{v}{u} \end{matrix} ) .

By taking (h̄^{2}, h̄^{3}, h̄^{4}) = (u_{1}, uv, 0), we have the following integrable system:

\{ \begin{matrix}u_{t} & = & u_{3} + \frac{3}{2}u_{1}u^{2} − 3u_{1}v^{2} − 3uvv_{1} \\ v_{t} & = & 3\frac{vu_{3}}{u} + v_{3} + 3uu_{1}v + \frac{3}{2}u^{2}v_{1} − 3v^{2}v_{1} + 3\frac{u_{1}v_{2}}{u} + 6\frac{u_{2}v_{1}}{u} \\ & & −3\frac{u_{1}u_{2}v}{u^{2}} − 3\frac{u_{1}^{2}v_{1}}{u^{2}} − 3vww_{1} − 3v_{1}w^{2} − 3\frac{u_{1}vw^{2}}{u} \\ w_{t} & = & 6\frac{wu_{3}}{u} + 3\frac{wv_{3}}{v} + w_{3} + 3uu_{1}w + \frac{3}{2}u^{2}w_{1} − 3w^{2}w_{1} \\ & & +9\frac{u_{2}w_{1}}{u} + 3\frac{u_{1}w_{2}}{u} + 6\frac{v_{2}w_{1}}{v} + 3\frac{v_{1}w_{2}}{v} + 3\frac{u_{1}v^{2}w}{u} \\ & & −12\frac{u_{1}u_{2}w}{u^{2}} − 3\frac{v_{1}v_{2}w}{v^{2}} + 6\frac{u_{2}v_{1}w}{uv} + 6\frac{u_{1}v_{2}w}{uv} + 6\frac{u_{1}v_{1}w_{1}}{uv} \\ & & −6\frac{u_{1}^{2}w_{1}}{u^{2}} − 3\frac{v_{1}^{2}w_{1}}{v^{2}} + 6\frac{u_{1}^{3}w}{u^{3}} − 6\frac{u_{1}^{2}v_{1}w}{u^{2}v} − 6\frac{u_{1}v_{1}^{2}w}{uv^{2}} . \end{matrix}  (23)

First let us write out Hasimoto transformation for n = 4 using Theorem 1.

v = \frac{cos θ_{24}}{cos θ_{34}} D_{x}θ_{23},

w = A_{3} = D_{x}θ_{34} + sin θ_{24}D_{x}θ_{23},

D_{x}θ_{24} = \frac{cos θ_{24}}{cos θ_{34}} sin θ_{34}D_{x}θ_{23},

which, together with the Euler transformation

u^{(1)} = u cos θ_{23} cos θ_{24}, u^{(2)} = u sin θ_{23} cos θ_{24}, u^{(3)} = u sin θ_{24}, transform equation (23) into the vector mKdV, i.e., u_{t} = u_{3} + \frac{3}{2}⟨u, u⟩u_{1}, where u = (u^{(1)}, u^{(2)}, u^{(3)})^{⊺}. The matrix T equals

cos θ_{23} cos θ_{24} sin θ_{23} cos θ_{24} sin θ_{24}
− sin θ_{34} sin θ_{24} cos θ_{23} − sin θ_{23} cos θ_{34} − sin θ_{34} sin θ_{24} sin θ_{23} + cos θ_{23} cos θ_{34} sin θ_{34} cos θ_{24}
− cos θ_{34} sin θ_{24} cos θ_{23} + sin θ_{23} sin θ_{34} − cos θ_{34} sin θ_{24} sin θ_{23} − cos θ_{23} sin θ_{34} cos θ_{34} cos θ_{24}

with the property T T^{⊺} = I and

D_{x}T = ( \begin{matrix}0 & v & 0 \\ −v & 0 & w \\ 0 & −w & 0 \end{matrix} ) T = \underline{ω}T

Now we compute the operator Q in Theorem 3

Q = T HT^{⊺}Φ̄_{2}^{∗}
= D_{x}Φ̄_{2}^{∗} − \underline{ω}Φ̄_{2}^{∗} + ∑_{i=1}^{2} ∑_{j=2}^{3} T J_{ij}UD_{x}^{−1}(J_{ij}U)^{⊺}T^{⊺}Φ̄_{2}^{∗}

= ( \begin{matrix}D_{x} & \frac{v}{u}D_{x} & \frac{w}{u} D_{x} \\ v & −D_{x} \frac{1}{u}D_{x} − u + \frac{w^{2}}{u} & −D_{x} \frac{w}{uv} D_{x} − \frac{w}{u} D_{x} \frac{1}{v} D_{x} − \frac{vw}{u} \\ 0 & −\frac{w}{u} D_{x} − D_{x} \frac{w}{u} & \frac{u}{v}D_{x} − \frac{w^{2}}{uv} D_{x} + D_{x} \frac{1}{u}D_{x} \frac{1}{v} D_{x} + D_{x} \frac{v}{u} \end{matrix} ) .

Equation (23) is bi-Hamiltonian with Hamiltonian function \frac{1}{2}(\frac{1}{4}u^{4} − u_{1}^{2} − u^{2}v^{2}) corresponding to Hamiltonian operator Φ̄_{2}Q and Hamiltonian function \frac{1}{2}u^{2} corresponding to Hamiltonian operator Φ̄_{1}Q.

According to the order of D_{x}, we can write Φ̄_{2}Q as the sum of three operators, namely,

H_{1} =

( \begin{matrix}D_{x} & \frac{v}{u}D_{x} & \frac{w}{u} D_{x} \\ D_{x} \frac{v}{u} & −D_{x} + 3\frac{w}{u} D_{x} \frac{w}{u} & −\frac{w}{v} D_{x} + \frac{w^{3}}{u^{2}v} D_{x} − D_{x} \frac{vw}{u^{2}} − \frac{w}{u} D_{x} \frac{v}{u} \\ D_{x} \frac{w}{u} & −D_{x} \frac{w}{v} + D_{x} \frac{w^{3}}{u^{2}v} − \frac{vw}{u^{2}} D_{x} − \frac{v}{u}D_{x} \frac{w}{u} & −2\frac{w}{u} D_{x} \frac{w}{u} + D_{x} + \frac{v}{u}D_{x} \frac{v}{u} \end{matrix} );

H_{2} = ( \begin{matrix}0 & 0 & 0 \\ 0 & −D_{x} \frac{1}{u}D_{x} \frac{1}{u}D_{x} & J_{23} \\ 0 & −J_{23}^{∗} & J_{33} \end{matrix} ) ,

where

J_{23} = −D_{x} \frac{1}{u}D_{x} \frac{w}{uv} D_{x} − D_{x} \frac{w}{u^{2}} D_{x} \frac{1}{v} D_{x} − \frac{w}{u} D_{x} \frac{1}{u}D_{x} \frac{1}{v} D_{x},

J_{33} = D_{x} \frac{1}{v} D_{x} \frac{1}{v} D_{x} − D_{x} \frac{1}{v} D_{x} \frac{w^{2}}{u^{2}v} D_{x} − D_{x} \frac{w^{2}}{u^{2}v} D_{x} \frac{1}{v} D_{x} − D_{x} \frac{w}{uv} D_{x} \frac{w}{uv} D_{x}

+D_{x} \frac{1}{v} D_{x} \frac{1}{u}D_{x} \frac{v}{u} + \frac{v}{u}D_{x} \frac{1}{u}D_{x} \frac{1}{v} D_{x}

and

H_{3} = ( \begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & D_{x} \frac{1}{v} D_{x} \frac{1}{u}D_{x} \frac{1}{u}D_{x} \frac{1}{v} D_{x} \end{matrix} ) .

Surprisingly, for any constants a and b, the operator a^{2}H_{1} + abH_{2} + b^{2}H_{3} is Hamiltonian.

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